AP Precalculus Exponential and Logarithmic Functions — Worked Answer Explanations

Unit 2 · 12 questions explained

1. Question 1 · Easy

   What is the next term of the geometric sequence $3,6,12,24,\dots$?

   • A
     $30$

     Why not A: Added a constant difference of $6$ — that would be arithmetic, not geometric.
   • B
     $36$

     Why not B: Multiplied $24$ by $1.5$ instead of the common ratio.
   • C
     $48$Correct
   • D
     $120$

     Why not D: Multiplied by $5$ instead of the common ratio $2$.
   Explanation

   Each term is obtained by multiplying the previous term by the common ratio $r=2$ (since $6/3=2$, $12/6=2$, $24/12=2$). The next term is $24\cdot 2=48$.

   Key takeaway

   Geometric sequences have a constant common ratio; the next term equals the previous term times $r$.

2. Question 2 · Easy

   Evaluate the logarithm ${\log }_{2}16$ without using a calculator.

   • A
     $2$

     Why not A: Confused the base with the answer.
   • B
     $4$Correct
   • C
     $8$

     Why not C: Computed $16/2$ instead of finding the exponent.
   • D
     $16$

     Why not D: Returned the argument rather than the exponent.
   Explanation

   ${\log }_{2}16$ asks: to what power must we raise $2$ to get $16$? Since $2^4=16$, the answer is $4$.

   Key takeaway

   $\log_b x = y$ means $b^y = x$ — the log is the exponent.

3. Question 3 · Easy

   If $f(x)=5\cdot 2^x$, what is $f(3)$?

   • A
     $30$

     Why not A: Multiplied $5\cdot 2\cdot 3$ — interpreted the exponent as a coefficient.
   • B
     $40$Correct
   • C
     $125$

     Why not C: Computed $5^3$ instead of $5\cdot 2^3$.
   • D
     $1000$

     Why not D: Computed $(5\cdot 2{)}^3$ — applied the exponent to the whole product, ignoring order of operations.
   Explanation

   Apply the exponent first by order of operations: $2^3=8$. Then multiply: $f(3)=5\cdot 8=40$.

   Key takeaway

   In $a \cdot b^x$, the exponent only acts on $b$ — multiply by $a$ after exponentiating.

4. Question 4 · Easy

   Solve ${\log }_3(x)=4$ for $x$.

   • A
     $x=7$

     Why not A: Added base and exponent instead of exponentiating.
   • B
     $x=12$

     Why not B: Multiplied base by exponent.
   • C
     $x=64$

     Why not C: Used base $4$ and exponent $3$ — reversed base and exponent.
   • D
     $x=81$Correct
   Explanation

   Rewrite in exponential form: ${\log }_3(x)=4$ means $3^4=x$. Compute $3^4=81$.

   Key takeaway

   $\log_b(x) = y \iff b^y = x$ — rewriting in exponential form is usually the fastest path.

5. Question 5 · Medium

   Simplify ${\log }_5(25)+{\log }_5(125)$ using log properties.

   • A
     $3$

     Why not A: Took only one of the two logs (${\log }_{5}125=3$) and ignored the other.
   • B
     $5$Correct
   • C
     ${\log }_5(150)$

     Why not C: Added the arguments — but the product rule says $\log a+\log b=\log (ab)$, not $\log (a+b)$.
   • D
     ${\log }_5(250)$

     Why not D: Doubled one term then added — algebra error.
   Explanation

   ${\log }_5(25)=2$ (since $5^2=25$) and ${\log }_5(125)=3$ (since $5^3=125$). Sum: $2+3=5$. Equivalently, ${\log }_5(25\cdot 125)={\log }_5(5^5)=5$.

   Key takeaway

   Product rule: $\log_b(xy) = \log_b(x) + \log_b(y)$.

6. Question 6 · Medium

   A bacterial culture grows according to $N(t)=500\cdot 2^{t/3}$ where $t$ is in hours. By what factor does the population grow every $3$ hours?

   • A
     Adds $500$.

     Why not A: Treated the function as linear with slope $500$.
   • B
     Doubles.Correct
   • C
     Triples.

     Why not C: Confused the base $2$ with the divisor $3$ in the exponent.
   • D
     Multiplies by $2^3=8$.

     Why not D: Used $t=3$ but ignored the $t/3$ in the exponent.
   Explanation

   Replace $t$ with $t+3$: $N(t+3)=500\cdot 2^{(t+3)/3}=500\cdot 2^{t/3}\cdot 2^1=2\cdot N(t)$. So the population doubles every $3$ hours.

   Key takeaway

   For $N(t) = a \cdot b^{t/k}$, the multiplier over a $k$-unit interval is exactly $b$.

7. Question 7 · Medium

   If $f(x)=e^x$ and $g(x)=\ln (x)$, what is $f(g(7))$?

   • A
     $0$

     Why not A: $f(0)=e^0=1$, not $0$ — also $g(7)\ne 0$.
   • B
     $1$

     Why not B: True for $g(1)=0$, but we are evaluating at $7$.
   • C
     $7$Correct
   • D
     $e^7$

     Why not D: Applied $f$ to $7$ instead of to $g(7)$ — skipped the inner function.
   Explanation

   $f$ and $g$ are inverse functions: $e^{\ln (x)}=x$ for $x>0$. So $f(g(7))=e^{\ln (7)}=7$.

   Key takeaway

   Exponential and logarithm with matching bases are inverses: $b^{\log_b x} = x$ and $\log_b(b^x) = x$.

8. Question 8 · Medium

   Solve $2^{x+1}=16$ for $x$.

   • A
     $x=2$

     Why not A: Likely set $x={\log }_2(16)-1=4-1=3$ then off-by-one.
   • B
     $x=3$Correct
   • C
     $x=4$

     Why not C: Took ${\log }_{2}16=4$ but ignored the $+1$ shift.
   • D
     $x=7$

     Why not D: Added base and result. Doesn't apply.
   Explanation

   Write $16=2^4$. Then $2^{x+1}=2^4$, so $x+1=4$ and $x=3$. Verify: $2^{3+1}=2^4=16$. ✓

   Key takeaway

   When bases match on both sides of an exponential equation, set the exponents equal.

9. Question 9 · Medium

   What is the inverse of $f(x)=3^x$?

   • A
     $f^{-1}(x)=\frac{1}{3^x}$

     Why not A: That is $3^{-x}$, the reciprocal, not the inverse function.
   • B
     $f^{-1}(x)={\log }_3(x)$Correct
   • C
     $f^{-1}(x)=x^{1/3}$

     Why not C: That inverts $x^3$, not $3^x$. Confused base and exponent positions.
   • D
     $f^{-1}(x)=\ln (x)$

     Why not D: $\ln$ inverts $e^x$, not $3^x$.
   Explanation

   Swap $x$ and $y$: $y=3^x\to x=3^y$. Solve for $y$ by writing in log form: $y={\log }_3(x)$. So $f^{-1}(x)={\log }_3(x)$.

   Key takeaway

   The inverse of $b^x$ is $\log_b(x)$ — base must match.

10. Question 10 · Hard

    Solve $\ln (x+2)-\ln (x)=\ln (3)$ for $x>0$.

    • A
      $x=-1$

      Why not A: Algebra leads here only if you forget the domain restriction $x>0$.
    • B
      $x=1$Correct
    • C
      $x=2$

      Why not C: Arithmetic error after applying the quotient rule.
    • D
      $x=3$

      Why not D: Set the inputs equal instead of using the quotient rule.
    Explanation

    Use the quotient rule: $\ln \left(\frac{x+2}{x}\right)=\ln (3)$, so $\frac{x+2}{x}=3$. Solve: $x+2=3x$, giving $2x=2$ and $x=1$. Check the domain: $x=1>0$, and $x+2=3>0$, so the solution is valid.

    Key takeaway

    Quotient rule: $\ln(a) - \ln(b) = \ln(a/b)$ — and always check the domain after exponentiating.

11. Question 11 · Hard

    A radioactive substance decays according to $A(t)=A_0\cdot (1/2{)}^{t/H}$ where $H$ is the half-life. If $A_0=80$ grams and $H=5$ days, how many grams remain after $15$ days?

    • A
      $5$ g

      Why not A: One extra half-life — used $t=20$ days instead of $15$.
    • B
      $10$ gCorrect
    • C
      $20$ g

      Why not C: Only halved once — used $t/H=1$ instead of $3$.
    • D
      $40$ g

      Why not D: Only halved once total — likely treated $H$ as the total decay time.
    Explanation

    $15/5=3$ half-lives. $A(15)=80\cdot (1/2{)}^3=80\cdot 1/8=10$ g.

    Key takeaway

    Each half-life multiplies the remaining amount by $1/2$ — count half-lives, not days.

12. Question 12 · Hard

    An investment of \2000$growscontinuouslyatanannualrateof$6%$.Accordingto$A(t) = 2000 \cdot e^{0.06 t}$,approximatelyhowlonguntilitdoubles?(Use$\ln 2 \approx 0.693$.)

    • A
      About $7.7$ years

      Why not A: Off by a factor — likely used $\ln 2/0.06-4$ or similar slip.
    • B
      About $11.6$ yearsCorrect
    • C
      About $16.7$ years

      Why not C: Used $1/0.06\approx 16.7$ — forgot the $\ln 2$ factor.
    • D
      About $33.3$ years

      Why not D: Used $2/0.06$ — multiplied by $2$ instead of taking $\ln 2$.
    Explanation

    Set $A(t)=2\cdot 2000=4000$: $4000=2000e^{0.06t}$, so $e^{0.06t}=2$, giving $0.06t=\ln 2\approx 0.693$. Therefore $t\approx 0.693/0.06\approx 11.55$ years.

    Key takeaway

    Doubling time under continuous growth: $t = \dfrac{\ln 2}{r}$ where $r$ is the continuous rate.